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Chapter 5: Problem 129

A mixture consisting of \(7.0 \mathrm{~g}\) of \(\mathrm{CO}\) and \(10.0 \mathrm{~g}\) of \(\mathrm{SO}_{2},\) two atmospheric pollutants, has a pressure of 0.33 atm when placed in a sealed container. What is the partial pressure of CO?

Short Answer

Expert verified
The partial pressure of \(\text{CO}\) is \0.203\text{ atm}\.

Step by step solution

01

- Calculate moles of \(\text{CO}\)

Use the molar mass of \(\text{CO}\) to convert grams to moles. The molar mass of \(\text{CO}\) is 28.01 g/mol. \Number of moles of \(\text{CO} = \frac{7.0 \text{ g}}{28.01 \text{ g/mol}} = 0.25 \text{ mol} \).
02

- Calculate moles of \(\text{SO}_{2}\)

Similarly, use the molar mass of \(\text{SO}_{2}\) to convert grams to moles. The molar mass of \(\text{SO}_{2}\) is 64.07 g/mol. \Number of moles of \(\text{SO}_{2} = \frac{10.0 \text{ g}}{64.07 \text{ g/mol}} = 0.156 \text{ mol} \).
03

- Total moles in the mixture

Add the moles of \(\text{CO}\) and \(\text{SO}_{2}\) to find the total moles in the mixture. \Total moles \(\text{n}_{\text{total}} = 0.25 \text{ mol} + 0.156 \text{ mol} = 0.406 \text{ mol} \).
04

- Calculate mole fraction of \(\text{CO}\)

The mole fraction \(\text{X}_{\text{CO}}\) is the ratio of moles of \(\text{CO}\) to the total moles. \Mole fraction \(\text{X}_{\text{CO}} = \frac{0.25 \text{ mol}}{0.406 \text{ mol}} = 0.616\).
05

- Calculate partial pressure of \(\text{CO}\)

The partial pressure of \(\text{CO}\) is found by multiplying its mole fraction by the total pressure. \Partial pressure of \(\text{CO} = \text{X}_{\text{CO}} \times P_{\text{total}} = 0.616 \times 0.33 \text{ atm} = 0.203 \text{ atm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Molar mass is an important concept in chemistry, particularly when dealing with gas calculations like partial pressure. It represents the weight of one mole of a given substance. The unit for molar mass is grams per mole (g/mol). To find the molar mass, you sum the atomic masses of all the atoms in a molecule.
Molar mass gives you a way to convert between grams and moles. For example, the molar mass of carbon monoxide (CO) is calculated by adding the atomic masses of carbon (12.01 g/mol) and oxygen (16.00 g/mol) to get 28.01 g/mol.
Knowing how to calculate and use molar mass is crucial for converting grams to moles, which is needed for calculating the number of molecules involved in a reaction or process. This understanding sets the foundation for further calculations involving mole fraction and partial pressure.
Mole Fraction
Mole fraction is a way of expressing the concentration of a component in a mixture. It is the ratio of the number of moles of one component to the total number of moles in the mixture.
To find the mole fraction \(X_{\text{CO}}\) of carbon monoxide (CO) in a mixture, follow this formula:
\[ X_{\text{CO}} = \frac{\text{moles of CO}}{\text{total moles}} \]
In our example, if you have 0.25 moles of CO and the total moles in the mixture is 0.406, then the mole fraction is:
\[ X_{\text{CO}} = \frac{0.25}{0.406} = 0.616 \]
Mole fraction is a dimensionless quantity and is essential in calculating partial pressures of gases in a mixture.
Partial Pressure
Partial pressure describes the pressure exerted by a single type of gas in a mixture of gases. According to Dalton's Law, the total pressure exerted by a gas mixture is equal to the sum of the partial pressures of each individual gas.
To calculate the partial pressure of carbon monoxide (CO) in our example, we use its mole fraction and the total pressure of the gas mixture. The formula is:
\[ P_{\text{CO}} = X_{\text{CO}} \times P_{\text{total}} \]
Given that the mole fraction of CO is 0.616 and the total pressure is 0.33 atm, the partial pressure of CO becomes:
\[ P_{\text{CO}} = 0.616 \times 0.33 \text{ atm} = 0.203 \text{ atm} \]
Partial pressures are useful in many chemical processes and reactions, particularly those involving gases.

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Most popular questions from this chapter

A gaseous organic compound containing only carbon, hydrogen, and nitrogen is burned in oxygen gas, and the volume of each reactant and product is measured under the same conditions of temperature and pressure. Reaction of four volumes of the compound produces 4 volumes of \(\mathrm{CO}_{2}, 2\) volumes of \(\mathrm{N}_{2},\) and 10 volumes of water vapor. (a) How many volumes of \(\mathrm{O}_{2}\) were required? (b) What is the empirical formula of the compound?

Aqueous sulfurous acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{3}\right)\) was made by dissolving \(0.200 \mathrm{~L}\) of sulfur dioxide gas at \(19^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg}\) in water to yield \(500.0 \mathrm{~mL}\) of solution. The acid solution required \(10.0 \mathrm{~mL}\) of sodium hydroxide solution to reach the titration end point. What was the molarity of the sodium hydroxide solution?

Aluminum chloride is easily vaporized above \(180^{\circ} \mathrm{C}\). The gas escapes through a pinhole 0.122 times as fast as helium at the same conditions of temperature and pressure in the same apparatus. What is the molecular formula of aluminum chloride gas?

A mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{Kr}\) weighs \(35.0 \mathrm{~g}\) and exerts a pressure of 0.708 atm in its container. since \(\mathrm{Kr}\) is expensive, you wish to recover it from the mixture. After the \(\mathrm{CO}_{2}\) is completely removed by absorption with \(\mathrm{NaOH}(s),\) the pressure in the container is 0.250 atm. How many grams of \(\mathrm{CO}_{2}\) were originally present? How many grams of Kr can you recover?

Convert each of the pressures described below to atm: (a) At the peak of Mt. Everest, atmospheric pressure is only \(2.75 \times 10^{2} \mathrm{mmHg}\) (b) A cyclist fills her bike tires to 86 psi. (c) The surface of Venus has an atmospheric pressure of \(9.15 \times 10^{6} \mathrm{~Pa}\) (d) At \(100 \mathrm{ft}\) below sea level, a scuba diver experiences a pressure of \(2.54 \times 10^{4}\) torr.
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